10x2+41x-41=0

Solution for 10x2+41x-41=0 equation:

10x^2+41x-41=0

a = 10; b = 41; c = -41;
Δ = b2-4ac
Δ = 412-4·10·(-41)
Δ = 3321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3321}=\sqrt{81*41}=\sqrt{81}*\sqrt{41}=9\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-9\sqrt{41}}{2*10}=\frac{-41-9\sqrt{41}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+9\sqrt{41}}{2*10}=\frac{-41+9\sqrt{41}}{20}
$